Source: encode_duration.go in package google.golang.org/grpc/internal/grpcutil

* * Copyright 2020 gRPC authors. * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http:www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. *


package grpcutil

import (
	"strconv"
	"time"
)

const maxTimeoutValue int64 = 100000000 - 1

div does integer division and round-up the result. Note that this is equivalent to (d+r-1)/r but has less chance to overflow.

func div(d, r time.Duration) int64 {
	if d%r > 0 {
		return int64(d/r + 1)
	}
	return int64(d / r)
}

EncodeDuration encodes the duration to the format grpc-timeout header accepts. https://github.com/grpc/grpc/blob/master/doc/PROTOCOL-HTTP2.md#requests

TODO: This is simplistic and not bandwidth efficient. Improve it.

	if t <= 0 {
		return "0n"
	}
	if d := div(t, time.Nanosecond); d <= maxTimeoutValue {
		return strconv.FormatInt(d, 10) + "n"
	}
	if d := div(t, time.Microsecond); d <= maxTimeoutValue {
		return strconv.FormatInt(d, 10) + "u"
	}
	if d := div(t, time.Millisecond); d <= maxTimeoutValue {
		return strconv.FormatInt(d, 10) + "m"
	}
	if d := div(t, time.Second); d <= maxTimeoutValue {
		return strconv.FormatInt(d, 10) + "S"
	}
	if d := div(t, time.Minute); d <= maxTimeoutValue {
		return strconv.FormatInt(d, 10) + "M"

Note that maxTimeoutValue * time.Hour > MaxInt64.

	return strconv.FormatInt(div(t, time.Hour), 10) + "H"